ASTIG M.E.

A. STEAM CYCLE DEFINITIONS: 1. Rankine cycle is the ideal cycle for vapor power plants. 2. In a rankine cycle, water enters the pump as saturated liquid and is compressed isentropically to the operating pressure of the boiler. 3. In the pump, the water pressure and temperature increases somewhat during this isentropic compression process due to a slight decrease in specific volume of water. 4. The superheated vapor enters the turbine and expands isentropically and produces work by the rotating shaft. The temperature and pressure may drop during the process. 5. steam is condensed at constant pressure in the condenser. 6. the boiler and condenser do not involve any work, and the pump and turbine are assumed to be isentropic. 7. Rankine cycle power plant converts 26% of the heat it receives in the boiler to net work. Work input 8. Back work ratio = Work output 9. The lesser the back work ratio, the better is the cycle. 10. Only 0.4 % of the turbine work output is required to operate the pump. 11. In actual condensers, the liquid is usually subcooled to prevent cavitation. That damaged the impeller of the pump. 12. Fluid friction causes pressure drops in the boiler, the condenser, and piping between various components. 13. The pressure in the condenser is usually very small. 14. To compensate pressure drops in rankine cycle, the water must be pumped to a sufficient higher pressure than ideal cycle. 15. The major source of irreversibility is the heat loss from the steam to the surroundings. 16. To increase the thermal efficiency of rankine cycle, increase the average temperature at which heat is transferred to the working fluid in the boiler. 17. To increase the thermal efficiency of rankine cycle, decrease the average temperature at which heat is rejected from the working fluid in the condenser. 18. Lowering the operating pressure of the condenser automatically lower the temperature of the steam, and thus the temperature at which heat is rejected. 19. The overall effect of lowering the condenser pressure is an increase in the efficiency of rankine cycle. 20. To take advantage of the increase efficiencies at low pressures, the condenser of steam power plants usually operate well below the atmospheric pressure. 21. The average temperature at which heat is added to the steam can be increased without increasing the boiler pressure by superheating the steam to high temperature.

22. Superheating the steam to higher temperature decreases the moisture content of the steam at the turbine exit. 23. Presently the highest steam temperature allowed at the turbine inlet is about 620 degree C. ceramics are very promising in this regard. 24. Raises the average temperature at which heat is added to the steam raises the thermal efficiency of the cycle. 25. The average temperature during the reheat process can be increased by increasing the number of expansion and reheat stages. 26. As the number of stages is increased, the expansion and reheat process approached an isothermal process at the maximum temperature. 27. In a reheat cycle, the optimum reheat pressure is about ¼ of the maximum cycle pressure. 28. The main purpose of reheating is to reduce the moisture content of the steam at the final stage of expansion. 29. Regeneration also provides a convenient means of dearating the feedwater to prevent corrosion in the boiler. 30. The cycle efficiency increases further as the number of feedwater heater is increased. 31. A trap allows the liquid to be throttled to a lower pressure region but traps the vapor. 32. A closed feedwater heater is more expensive than open feedwater heater. 33. Cogeneration is the production of more than one useful of energy (such as process heat and electric power)from the same energy source. 34. The overall thermal efficiency of a power plant can be increased by binary cycles or combined cycles. 35. A binary cycle is composed of two separate cycles, one at high temperature (topping cycle) and the other at relative low temperature. 36. Combined cycles have a higher thermal efficiency than the steam or gas turbine cycle operating alone. METHODS OF IMPROVING THE EFFICIENCY OF RANKINE CYCLE 1. By lowering the condenser pressure in rankine cycle (1) (4) (4’) T

(3)

(2)

(3’) (2’) Increase in W

S

In a rankine cycle with fixed turbine inlet condition. What is the effect of lowering the condenser pressure? A. The pump work input will increase. B. The turbine work output will increase C. The heat added will increase. D. The heat rejected will decrease. E. The cycle efficiency will increase. F. The moisture content at turbine exit will increase. 2. By increasing the boiler pressure in rankine cycle (1’) (1)

T1 = T2

increase in W decrease in W (4’) T (4) (3’) (2’) (2)

S In an ideal Rankine cycle with fixed turbine inlet temperature and condenser pressure, what is the effect of increasing boiler pressure. A. The pump work input will increase . B. The turbine work output will increase . C. The heat added will decrease. D. The heat rejected will decrease. E. The cycle efficiency will increase. F. The moisture content at turbine exit will increase 3. By superheating the steam to higher temperature in rankine cycle (1’) (1) (4)

increase in W

T (3) (2) (2’)

S

In an ideal rankine cycle with fixed boiler and condenser pressures, what is the effect of superheating the steam to a higher temperature. A. The pump work input will remains constant. B. The turbine work output will increase C. The heat added will increase D. The heat rejected will increase E. The cycle efficiency will increase F. The moisture content at turbine exit will decrease 4. By reheating the steam in rankine cycle. (1)

(3) (2)

(6)

REHEATING

T (5) (4)

S Assume the mass flow rate is maintained the same, when a simple ideal rankine cycle is modified with reheating, A. The pumpwork input will remains constant. B. The turbine work output will increase C. The heat added will increase. D. The heat rejected will increase. E. The cycle efficiency will increase. F. The moisture content at turbine exit will decrease. 5. Regeneration of the steam in rankine cycle How do the following quantities change when the simple ideal rankine cycle is modified with Regeneration? A. The pump work input will decrease. (1)

1 (2) (1 – m) (1 – m)

(3)

B. The turbine work output will decrease C. The heat added will decrease D. The heat rejected will decrease E. The cycle efficiency will increase F. The moisture content at turbine exit remains constant 1. RANKINE CYCLE Is the standard and most common steam cycle.the working fluid is water SCHEMATIC DIAGRAM:

QA (1) (4)

TURBINE

BOILER WT

WP PUMP (2) CONDENSER QR

(3) (1’) P4 = P1 (4)

S1 = S2

T S3= S4 (3) P3 = P2

(2)

S FORMULAS QA 1.1 Heat addition in boiler, QA m

BOILER

(1)

The Boiler conditions: 1. Changes in kinetic and potential energy are negligible 2. There is no work crossing the control surface 3. The process is constant pressure heat addition QA = ( h1-h4 ), kJ/kg QA= m( h1-h4 ), kw Where: m = mass of steam flow 1.2 Turbine work, Wt m (1) WT Turbine (2) The turbine conditions: 1. The turbine is adiabatic Q = 0 and s1 = s2 2. The change in kinetic energy across the turbine is not negligible but in the ideal analysis the kinetic energy change is being ignored since we have no specific means of determining inlet and discharges velocities. W = h1-h2, kJ/kg

W = m ( h1-h2 ), kW

1.3 Quality after turbine expansion, X

s = s = sf + x sfg 2

1

S1 - Sf

2

X2 = Sfg h2 = hf2 + x2hfg2 1.4 Heat rejected in the condenser, QR The condenser conditions is The same as the boiler conditions QR = h2 - h3, kJ/kg QR = m (h2 - h3), kW m (2) Cooling requirement in condenser QR = Q mw, mass flow of cooling water

T2

WATER

CONDENSER

QR

m (h2 - h3) mw =

(3) CP (t2 - t1)

mw t1 COOLING WATER

1.4 Pump work, WP

(4) WP PUMP (3)

Pump conditions: 1. The pump is adiabatic (Q= 0 ) And reversible (s1 = s2) 2. The change in kinetic and potential energy are negligible 3. The fluid is not in compressible (v1 = v2) WP = h4-h3, kJ/kg Wp = v3 (p -p3),kJ/kg WP = m (h4-h3),kW

Enthalpy calculations at point 4

4

h4 = v3(P4-P )+h3 3

1.6 Cycle Efficiency, e turbine work – pump work e

= heat added WT - WP e =

(h1 - h2) - ( h4 - h3 ) =

QA 1.7 Cogeneration efficiency ec

(h1-h4)

( QR + WT ) ec = QA 1.8 Backwork ratio (BW) W h4 - h3 BW = = W h1 - h2 P

T

Problems 1. In a rankine cycle steam enters the turbine at 2.5Mpa (enthalpies and entropies given) and condenser of 50kpa (properties given), what is the thermal efficiency of the cycle? Ans. 25.55% 2. In an ideal rankine cycle the steam throttle condition is 4.10Mpa and 440 C. if turbine exhausts is 0.105Mpa ,determine the thermal efficiency of the cycle Ans.

27.55%

3. In a rankine cycle saturated liquid water at 1 bar is compressed isentropically to 150 bar.First by heating in a boiler,and then by super heating at constant pressure of 150 bar. The water substance is brought to 750 K. After adiabatic reversible expansion in a turbine To 1 bar,it is then cooled in a condenser to a saturated liquid. What is the thermal efficiency of the cycle? Ans.

34.24% 2. REHEAT CYCLE

Schematic diagram: Turbine WT

(1) HP BOILER

QA

(2)

LP (3) (4)

condenser QR Pump (6)

(5)

WP REHEATER

(1) (3) P3 = P2

S1 = S2

P3 = P2

T

(6) (5)

(2)

S3 = S4

P5 = P4 (4)

S

c (6)

2.2 TURBINE WORK WT = (h1 – h2) + (h3 – h4), Kj/Kg WT = ms[(h1 – h2) + (h3 – h4)],kw

(1)

HP

LP

WT

(2) (3)

2.3 PUMP WORK WP = h6 – h5, kj/kg WP = ms(h6 – h5), kw WP = V5(P6 – P5), kj/kg

(4)

(6) (5)

PUMP WP

2.4 ENTHALPY h6 = V5(P6 – P5 )+ h5 2.5 HEAT REJECTED IN THE CONDENSER QR = ( h4 – h5 ), kj/kg QR = ms ( h4 – h5 ), kj/kg

mw

(4)

t2 COOLING WATER REQUIREMENT IN CONDENSER QR = QWATER Mw = mass flow of cooling water

QR

ms(h4 – h5), MW = CP (T2 – T1) 2.6 THERMAL EFFICIENCY

mw (5)

t1

cooling water

( WT – WP) e= QA PROBLEM S 1. Steam is delivered to turbine at 5.4 Mpa and 600C.before condensation at 31C, steam is extracted for feedwater heating at 0.6Mpa. for an ideal regenerative cycle, find the thermal efficiency. Ans.

44.14 %

2. Steam enters the turbine of a cogeneration plant at 7.0Mpa and 500C .Steam at a flow rate of 7.6 kg/sec is extracted from the turbine at 600kpa pressure for process heating. The remaining steam continues to expand 10kpa. The recovered condensates are pumped back to the boiler. The mass flow rate of the steam that enters the turbine is 30 kg/s. calculate the cogeneration efficiency percent . Ans.

79.81% 3. REGENERATIVE CYCLE

SCHEMATIC DIAGRAM 1 kg WT

(1)

Turbine QA

BOILER

(2) (2)

(3) (1 - m)

m

condenser

O.H

QR 1 kg

(5) (6) (7)

pump 1

(4) (1 – m) pump 2

(1) T

1

1 P7=P1 m P6=P2 (1 – m)

(3)

FORMULAS 3.1 Heat Added (1) BOILER QA

(7) QA=ms(h1-h7

TURBINE

3.2 Turbine Work WT=1(h1-h2) + (1-m)(h2-h3) m = mass of the extracted steam 1 = mass of supply steam 3.3 Pump Work WP = Wp1 + Wp2 WP = (1 – m)(h5-h4)+1(h7-h6) 3.4 Enthalpy calculation at point 5and 7

OPEN FWH

h5 =v4(p5-p4)+h4 h7 =v6(p7-p6)+h6 3.5 Heat rejected QR=(1-m)(h3-h4) WT-WP 3.6 Efficiency = QA 3.7 Heat balance in regenerative heater mh2+(1-m)h5=mh6 where: m=mass of extracted steam (h6-h5) m= (h2-h5)

OPEN FWH

Problem 1. A turbine with one extraction for regenerative feedwater heating receives steam with an enthalpy of 3373 kJ/kg and discharges it with an exhaust enthalpy of 2326 kJ/kg. the ideal regenerative feedwater heater receives 11338 kg/hr of extracted steam at 345 kpaa (whose h=2745 kJ/kg ). The feedwater (condensate from the condenser ) enters the heater with an enthalpy of 140 kJ/kg and departs saturated at 345 kpaa (hf=582 kJ/kg ). Calculate the turbine power. Ans:

18114 kW

2. There are receive 150000ib/hr of steam by an ideal regenerative engine, having only one heater, of which the heater receives 33950 ib/hr the condenser receives the remainder at 1 psia if the heater pressure is 140 psia find the state (quality or sh ) of the steam a) at the entrance. Ans:

sh=3.64 F

4. CARNOT CYCLE -Is the most efficient thermodynamic steam cycle. FORMULAS: 4.1 Heat added QA = T1(S1-S4)

QA

4.2 Heat rejected QR = T2(S2 - S3) QR = T2(S1-S4) 4.3 Work output W = QA – QR W = T1(S1-S4) - T2(S1-S4)

T=c

T S=c

4.4 Cycle efficiency W QA – QR W e= = = QA QA W + QR In terms of temperature: T2 – T1 e= T1 Where: T1 = temperature source, abs T2 = temperature sink, abs

QR ENTROPY,S

PROBLEM 1. A carnot engine operating between 900F and 90F produces 40000 ft-lb of work. Determine (a) the heat supplied (b) the change of entropy during heat rejection , and (c) thermal efficiency of the engine. Ans. (a) 86.3 btu (b) 0.0635 btu/R (c) 59.56% 2. A reverse carnot cycle requires 3HP and extracts energy from a lake to heat a house, if a house is kept at 70F and requires 2000Btu/min, what is the temperature of the lake? Ans. 36 F B. STEAM ENGINES DEFINITIONS: 1. Steam engine – Is a heat engine supplied by steam which presses the piston on one side and followed on the other side that cause to move the cylinder in reciprocating motion which converted into mechanical power or shaft power. 2. Single acting engine – Is one which work is done on the one side of the piston having only one working stroke to a revolution. 3. Engine indicator – Traces the actual PV – diagram. 4. Double acting engine – Is an engine in which work is done on both sides of the piston. 5. Planimeter – Is an instrument use to measure the area of PV diagram. 6. Dynamometer – measures the torque. 7. Tachometer – measures the speed of the engine. 8. Break horse power – The actual horse power delivered by their engine to the driven shaft. 9. Indicated power – The power developed in the engine cylinder as obtain from the pressure in the cylinder. 10. Friction power – the pressure and torque spent in over coming friction of reciprocating and revolving parts of the engines and the automobile before it reached drive shaft. 11. Combine horse power – The final horse power delivered by the engine to the drive shaft. 12. Mean effective pressure – Is the average pressure acting the piston during the operating cycle.

Steam h1 ms

HEAD END

CRANK END d

D

INDICATED POWER h2

t2 condenser QR mw

t1

hf2 cooling water FORMULAS:

1.Volume displacement, Vd a) Piston rod Neglected: Note: Steam engine is a double acting engine.

P

2

2 ¶( D) L N 4 V b) Piston rod considered: 2

VD 2

2

3

2 ¶( D) L N + ¶( D - d ) L N , m / sec 4 4 Where; d = piston rod diameter 2. Indicator diagram is used to find the mean indicated pressure. a) if average area is given: Pmi = A K

P

Pmi

L Where: A= average area of diagram for both sides of piston L= length of the diagram K= spring scale constant, kpa/m or psi/in Pmi = indicated mean effective pressure b) if two areas are given for both sides of the piston: A1+A2 2 K L 3. Indicated power, IP IP = Pmi x VD

V

where: Pmi =indicated power

4. Brake power( BP ) BP = Pmb x VD, kW Bp = 2 ¶ T N, kW

where:

Pmb = Brake mean effective pressure T = torque , kN. M N = speed , rps

5. Friction power ,FP FP = Indicated power 6. Mechanical efficiency, nm BP nm = IP 7. Indicated thermal efficiency, nti

IP nti = ms(h1 – h2)

8. Brake thermal efficiency, ntb BP ntb = ms(h1 – h2) 9. Indicated engine efficiency,

nei IP nei = ms(h1 – h2)

10. Brake enginel efficiency, neb BP ntb =

ms(h1 – h2 ) 11. Cooling water requirement in condenser ms(h1 – h2 ) QR = QWATER mw = mass flow of cooling water

mw = Cp(t2 – t1)

PROBLEM 1. A 254 mm X 305 mm steam engine has 45 mm diameter piston rod. indicator card are as follows : Head end card …………850 sq.mm Crank end card ……………..930 sq.mm Length of card…………. 74 mm Spring scale…………….21.75 kpa/mm Breakarm………………..1.5 m Net brake weight………...780 N Angular speed……………4.2 rev/sec Calculate: a. indicated power b. brake power c. friction power d. mechanical efficiency ans. a. 33.394 Kw b. 30.87557 Kw c. 2.51843 Kw d. 92.458% 2.. The crank shaft double acting steam engine rotates at 220rpm the bore and stroke of the steam engine is 300mm X 450mm, and the mean effective pressure acting upon the piston is 4kg/sq.cm. find the indicated horse power or the horse power developed in the cylinder. Ans. 122.655 Hp

3. A 350mm X 450mm engine running at 260 rpm has an entrance steam condition of 2Mpa and 230 C and the exit at 0.1 Mpa the steam consumption is 2000 kg/hr and mechanical eff. Is 88% if indicated mean effective pressure is 600Kpa det. Indicated thermal eff. Ans. 16.66%

C.STEAM TURBINE DEFINITIONS: 1. Steam turbine - is a heat engine in which the potential, kinetic energy and enthalpy of a steam is change into useful work. 2. Brake horse power – the actual horse power delivered by their engine to the driven shaft. 3. Indicated power – the power developed in the engine cylinder as obtained from the pressure in the cylinder. 4. Friction power – the pressure and torque spent in over coming friction of reciprocating and revolving parts of the engine and the automobile before it reach the drive shaft. 5. Combined horse power – The final horse power delivered by the engine to the drive shaft. 6. Mean effective pressure – Is the average pressure acting the piston during the operating cycle. TWO BASIC TYPES OF STEAM TURBINES 1. Impulse steam turbine – the steam expands in the nozzle, losing its pressure but gaining velocity. 2. Reaction steam turbine – The steam expands in the blading. PURPOSE OF DUMMY PISTON IN STEAM TURBINE -Is to counter act pressure difference across the blading in the reaction turbines, which tends to removes the rotor axially towards the exhaust end. METHODS IN DETERMINING THE OUTPUT OF A TURBINE 1. by mechanical means such as prony brakes or water brakes. 2. Torsion meter kby measuring the output electrically through the dynamometer. 3. by a thrust meter. Steam ms

Internal power h1

GENERATOR

WT STEAM TURBINE

h2 GENERATOR OUTPUT

t2

QR CONDENSER

mw t1 hf2

COOLING WATER

FORMULAS: 1. If kinetic energy and heat loss is considered 2

2

W = ms(h1 – h2) + ½ ms ( v1 – v2 ) – Q

(1) Actual

2. Ideal turbine work:(KE and Q neglected) Ideal turbine work = ms(h1 –h2)

h

(2) (3)

3. Actual turbine work: Actual turbine work = ms(h1 –h2a) Actual turbine work = ms(h1 –h2a) nst where: h2a = enthalpy after actual expansion nst = stage efficiency 3. Stage efficiency, nst (h1 –h2a) nst = (h1 –h2) 5. Turbine power output: Turbine power output = ms(h1 –h2a) nT where: nT = turbine efficiency 6. Generator efficiency, nG Generator output nG = Turbine output 7. Brake thermal efficiency: ntb Turbine output ntb = ms(h1 –hf2) 8. Combined thermal efficiency: Generator output ntc= ms(h1 –hf2)

Ideal s1 = s2 entropy, s

9.

Brake engine efficiency: neb Brake power

neb = ms(h1 –h2) 10. Combined efficiency:, nec Generator output ntc = ms(h1 –h2) 11. Cooling water requirement in condenser QR = Qwater mw = mass flow of cooling ms(h1 – h2 ) mw = Cp(t2 – t1) 12. Willan’s line - Is a straight line which shows the relation between the steam consumption (ms, kg/hr ) and the load ( L, kw) of steam turbine generator unit.

ms,kg/hr

No load

load

full load

PROBLEM 1. A steam turbine receives 5000 kg/hr of steam at 5Mpa and 400C and velocity of 25 m/sec. it leaves the turbine at 0.006 Mpa and 85% quality and velocity of 20 m/sec. radiation loss is 10000 kj/hr. find the kw developed. Ans. 1373.29 kw

2. A steam turbine with 90% stage efficiency receives steam at 7 Mpa and 550C and exhaust as 20 kpa. Determine the turbine work? Ans.

117.01 kj/kg

3. A steam turbine power plant of 5,000 kw capacity has a full load steam rate of 6 kg steam per kw-hr . no load steam consumption may be taken as 10% of the full load steam consumption.write the equation of willans line of this turbine and at 60% of rate load, calculate the hourly steam consumption of this unit. Ans.19200 kg/hr

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STEAM POWER PLANT • • • • • • •

Steam Cycle Rankine Cycle Reheat Cycle Regenerative Cycle Carnot Cycle Steam Engine Steam Turbine